We will make use of the fact that if #P(x)# is a polynomial, then #(x-a)# is a factor of #P(x)# if and only if #P(a) = 0#, along with polynomial long division.

If we look for a root of #x^4-2x^3+2x-1# (that is, a value such that the expression evaluates to #0#), we find that #1^4-2(1)^3+2(1)-1=0#. Thus, using the above fact, we know that #(x-1)# is a factor of #x^4-2x^3+2x-1#. Dividing, we find:

#(x^4-2x^3+2x-1)/(x-1) = x^3-x^2-x+1#

#=> x^4-2x^3+2x-1=(x-1)(x^3-x^2-x+1)#

Next, we will look to factor #x^3-x^2-x+1#. While there are more general ways of factoring cubic expressions, we can still test for roots first. Doing so, we find that #1# is still a root of the remaining cubic equation, and so we once again divide by #(x-1)#:

#(x^3-x^2-x+1)/(x-1) = x^2-1#

#=> (x^3-x^2-x+1) = (x-1)(x^2-1)#

For the quadratic equation #x^2-1#, there are multiple ways of continuing. The easiest is to note that it is an example of the special product #(a+b)(a-b) = a^2-b^2# with #a=x# and #b=1#. We could also test for roots as we did above and divide. Either way, we will find that

#x^2-1 = (x+1)(x-1)#

Now that we have factored everything down to polynomials of degree #1#, we can put it all together to get our factored form:

#x^4-2x^3+2x-1 = (x-1)(x^3-x^2-x+1)#

#=(x-1)(x-1)(x^2-1)#

#=(x-1)(x-1)(x+1)(x-1)#

#=(x-1)^3(x+1)#